KCSE Algebra Guide — Linear, Quadratic and Simultaneous Equations

Introduction to Algebra

Algebra is the branch of mathematics in which letters (called variables or unknowns) are used to represent numbers. Instead of asking "what number plus 5 equals 12?", algebra lets us write x + 5 = 12 and then solve for x systematically. This approach is enormously powerful — it lets us solve problems involving unknown quantities across every field from engineering to economics.

In KCSE Mathematics, algebra spans all four forms. This guide covers the three most important equation types: linear equations (Form 1–2), quadratic equations (Form 2–3) and simultaneous equations (Form 2–3).

Part 1 — Linear Equations

A linear equation is one in which the unknown (x) appears to the power of 1 only — no x², no x³. The graph of a linear equation is always a straight line. The general form is ax + b = c, where a, b and c are numbers.

Method: Isolate the Unknown

To solve a linear equation, perform the same operation on both sides until x is alone on one side. The key rule is: whatever you do to one side, do to the other.

Example 1 — Simple linear equation

Solve: 3x + 7 = 22

Subtract 7 from both sides: 3x + 7 − 7 = 22 − 7 → 3x = 15

Divide both sides by 3: x = 15 ÷ 3

x = 5

Check: 3(5) + 7 = 15 + 7 = 22 ✓

Example 2 — Variable on both sides

Solve: 5x + 3 = 2x + 15

Subtract 2x from both sides: 3x + 3 = 15

Subtract 3 from both sides: 3x = 12

Divide by 3: x = 4

x = 4

Check: 5(4)+3 = 23; 2(4)+15 = 23 ✓

Example 3 — Equation with fractions

Solve: x/3 + 2 = 5

Subtract 2 from both sides: x/3 = 3

Multiply both sides by 3: x = 9

x = 9

Part 2 — Quadratic Equations

A quadratic equation contains x² as the highest power. The general form is ax² + bx + c = 0. Quadratic equations can have two solutions, one solution or no real solutions. There are three main methods for solving them.

Method 1: Factorisation

Factorisation works when the quadratic can be written as a product of two brackets. It is the fastest method when it works.

To factorise ax² + bx + c, find two numbers that multiply to give ac and add to give b.

Example 4 — Factorise and solve

Solve: x² − 5x + 6 = 0

Find two numbers that multiply to 6 and add to −5: they are −2 and −3

Factorise: (x − 2)(x − 3) = 0

Set each factor to zero: x − 2 = 0 or x − 3 = 0

x = 2 or x = 3

Example 5 — Difference of two squares

Solve: x² − 16 = 0

Recognise as difference of squares: (x + 4)(x − 4) = 0

x + 4 = 0 or x − 4 = 0

x = −4 or x = 4

Method 2: The Quadratic Formula

The quadratic formula works for ANY quadratic equation, even when factorisation is not possible. For ax² + bx + c = 0:

x = (−b ± √(b² − 4ac)) / 2a

The expression b² − 4ac is called the discriminant. If it is negative, there are no real solutions.

Example 6 — Quadratic formula

Solve: 2x² + 3x − 2 = 0 (a=2, b=3, c=−2)

Discriminant: b² − 4ac = 9 − 4(2)(−2) = 9 + 16 = 25

x = (−3 ± √25) / (2×2) = (−3 ± 5) / 4

x = (−3 + 5)/4 = 2/4 = 1/2, or x = (−3 − 5)/4 = −8/4 = −2

x = 1/2 or x = −2

Method 3: Completing the Square

Completing the square rewrites the quadratic in the form (x + p)² = q, then takes the square root of both sides. It is useful when the coefficient of x² is 1.

Example 7 — Completing the square

Solve: x² + 6x + 5 = 0

Move constant: x² + 6x = −5

Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4

Write as square: (x + 3)² = 4

Take square root: x + 3 = ±2

x = −1 or x = −5

Part 3 — Simultaneous Equations

Simultaneous equations are two or more equations that share the same unknown values. The solution is the pair of values (x, y) that satisfies both equations at the same time. There are two main methods: elimination and substitution.

Method 1: Elimination

Add or subtract the equations to eliminate one variable, then solve for the other.

Example 8 — Elimination method

Solve: 3x + 2y = 16 and 5x − 2y = 8

Add the equations (the y terms cancel): 8x = 24 → x = 3

Substitute x = 3 into first equation: 9 + 2y = 16 → 2y = 7 → y = 3.5

x = 3, y = 3.5

Check in second equation: 5(3) − 2(3.5) = 15 − 7 = 8 ✓

Method 2: Substitution

Make one variable the subject of one equation, then substitute that expression into the other equation.

Example 9 — Substitution method

Solve: y = 2x − 1 and 3x + y = 14

Substitute y = 2x − 1 into the second equation: 3x + (2x − 1) = 14

Simplify: 5x − 1 = 14 → 5x = 15 → x = 3

Find y: y = 2(3) − 1 = 5

x = 3, y = 5

KCSE Examination Tips for Algebra

Always show your method clearly. In KCSE, method marks are awarded even if your final answer is wrong. Write every step.
Check your answers by substituting back into the original equation — this catches arithmetic errors before you submit.
For quadratic equations, try factorisation first. If you cannot factorise within 30 seconds, use the quadratic formula — it always works.
For simultaneous equations, choose elimination when the coefficients of one variable are equal or opposite. Choose substitution when one equation already gives y (or x) in terms of the other variable.
Watch signs carefully — the most common mistakes in algebra are sign errors when subtracting negative numbers.

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